+0

0
11
2
+15

Let x and y be nonnegative real numbers. If x^2+3y^2=18, then find the maximum value of xy.

Feb 16, 2024

#2
+2

As we need to find the max value of a product we can use AM-GM Inequality. the inequality states that for any real numbers $$x_1, x_2, \ldots, x_n \geq 0$$,
$$\LARGE\frac{x_1 + x_2 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \cdots x_n}$$

with equality if and only if $$x_1 = x_2 = \cdots = x_n$$

Thus using this we write:
$$\Large\frac{x^2+3y^2}{2}\ge\sqrt[2]{x^2\cdot3y^2}$$

Plugging in values and solving-->
$$\sqrt3xy\le9\\ \boxed{xy\le3\sqrt3}$$

Feb 16, 2024

#1
+53
+1

I guess draw the ellipse

Feb 16, 2024
#2
+2

As we need to find the max value of a product we can use AM-GM Inequality. the inequality states that for any real numbers $$x_1, x_2, \ldots, x_n \geq 0$$,
$$\LARGE\frac{x_1 + x_2 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \cdots x_n}$$

with equality if and only if $$x_1 = x_2 = \cdots = x_n$$

Thus using this we write:
$$\Large\frac{x^2+3y^2}{2}\ge\sqrt[2]{x^2\cdot3y^2}$$

Plugging in values and solving-->
$$\sqrt3xy\le9\\ \boxed{xy\le3\sqrt3}$$